# Totally bounded subsets gives separable set

Suppose $$X$$ is a metric space that is covered by a countable collection of totally bounded subsets of X. Show that X is separable.

My thoughts:

Let's say $$C_n$$ is the collection. If $$X$$ is covered by countable collection of totally bounded subsets of $$X$$ then the union $$\bigcup C_n$$ is countable and totally bounded. The union covers the set, $$X \subseteq \bigcup C_n$$ and so $$X$$ is totally bounded and thus separable.

Any help would be appreciated.

• The union is not countable but it is a countable union of separable sets hence separable. Saying "the union is countable" is misleading. – Henno Brandsma Dec 2 '18 at 16:09

For metric spaces being (hereditarily) separable, (hereditarily) Lindelöf, and second countable are all equivalent (e.g. see here) and a totally bounded subset has all of these properties (e.g. it's easy to see that totally bounded implies separable, by applying the definition to all $$\varepsilon$$ of the form $$\frac{1}{n}$$).
So if $$X$$ is covered by countably many totally bounded sets, it is also covered by countably many separable sets and is so separable itself (the union of countably many dense subsets is dense in the union). So $$X$$ is also separable.
$$X=\cup_{i\geq 1} X_i$$ where $$X_i$$'s are totally bounded. For fixed natural numbers $$i$$ and $$n$$ there are finitely many points $$x_{i1},x_{i2}\ldots x_{im_n}$$ in $$X_i$$ such that $$X_i\subset \cup_{j=1}^{m_n}B(x_{ij},1/n).$$ Then $$A_i=\cup_{n\geq 1}\{x_{i1},x_{i2}\ldots x_{im_n}\}$$ is a countable dense subset of $$X_i.$$ Hence $$A=\cup_{i\geq 1}A_i$$ is countable dense subset of $$X.$$